A) \[{{\log }_{e}}\left( 1+\frac{1}{x} \right)\]
B) \[{{\log }_{e}}\left( 1-\frac{1}{x} \right)\]
C) \[{{\log }_{e}}\left( \frac{x}{x+1} \right)\]
D) None of these
Correct Answer: A
Solution :
Given series is \[\frac{1}{x+1}+\frac{1}{2{{(x+1)}^{2}}}+\frac{1}{3{{(x+1)}^{3}}}+.......\infty \] \[=-{{\log }_{e}}\left( 1-\frac{1}{x+1} \right)=-{{\log }_{e}}\left( \frac{x}{x+1} \right)\] \[={{\log }_{e}}\left( \frac{x+1}{x} \right)={{\log }_{e}}\left( 1+\frac{1}{x} \right)\].You need to login to perform this action.
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